Question

A particle executes simple harmonic motion and is located at x= a, b and c at times ${t_0}$, 2${t_0}$ and 3${t_0}$ respectively. The frequency of the oscillation is
A. $\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + b}}{{2c}}} \right)$
B. $\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + b}}{{3c}}} \right)$
C. $\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{2a + 3c}}{b}} \right)$
D. $\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right)$

Answer 1

Hint: Use the given information and make equations then add equation $a = A\cos \omega {t_0}$ and $c = A\cos 3\omega {t_0}$ then simplify it and find the value of frequency of the oscillation.

Complete Step-by-Step solution:
According to the given information location of particles located at x are a, b and c at time ${t_0}$, 2${t_0}$, 3${t_0}$.
So the equations of the particles are
$a = A\cos \omega {t_0}$ ---------- (Equation 1)
$b = A\cos 2\omega {t_0}$ ------------- (Equation 2)
$c = A\cos 3\omega {t_0}$ --------------- (Equation 3)
On adding equation 1 and 3
$a + c = A\cos 3\omega {t_0} + A\cos \omega {t_0}$
$a + c = A(\cos 3\omega {t_0} + \cos \omega {t_0})$
Applying the formula $\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$
$a + c = 2A\cos (\dfrac{{3\omega {t_0} + \omega {t_0}}}{2})\cos (\dfrac{{3\omega {t_0} - \omega {t_0}}}{2})$
$a + c = 2A\cos 2\omega {t_0}\cos \omega {t_0}$
Form equation 2 $b = A\cos 2\omega {t_0}$
Therefore $a + c = 2b\cos \omega {t_0}$
Since $\omega = 2\pi f$
$ \Rightarrow $${\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) = 2\pi f{t_0}$
$ \Rightarrow $$f = \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right)$
Hence option D is the correct option.

Note: You can see that in the above solution we have used the term Simple Harmonic Motion which is defined as a state of a motion in which the restoring force is directly proportional to the displacement of the body from its mean position and the direction of this restoring force is always towards the mean position. For this state of motion the acceleration of a particle executing simple harmonic motion is given by, $a\left( t \right) = - {\omega ^2}x\left( t \right)$. Here, ω is the angular velocity of the particle.